# limitWatch

This discussion is closed.
#1
lim(x->0) (e^(-1/x^2))/x

thanks for any help
0
13 years ago
#2
(Original post by mik1w)
lim(x->0) (e^(-1/x^2))/x

thanks for any help
as x gets very small, -1/(x^2) gets 'very negative' - it tends towards "minus infinity"

e^'very negative' = 'very small'. as 'very negative' becomes "minus infinity", 'very small' becomes 0.
0
13 years ago
#3
as x gets very small, -1/(x^2) gets very negative - it tends towards 'minus infinity'

e^(very negative) = very small. as 'very negative' becomes 'minus infinity', 'very small' becomes 0.
i liked ur "expressions"
0
13 years ago
#4
(Original post by yazan_l)
i liked ur "expressions"
0
13 years ago
#5
(Original post by mik1w)
lim(x->0) (e^(-1/x^2))/x

thanks for any help
You can write it as (1/x).1/e^(1/x^2) which clearly shows it to be 0 as e^(1/x^2) will approach infinity much faster than (1/x) does.
I can't see how to prove it at a more advanced level.
0
13 years ago
#6
(Original post by El Stevo)
as x gets very small, -1/(x^2) gets 'very negative' - it tends towards "minus infinity"

e^'very negative' = 'very small'. as 'very negative' becomes "minus infinity", 'very small' becomes 0.
yes you're dividing by x which is also becoming very small
0
13 years ago
#7
The way I did it was using the substitution t=1/xÂ² to transform it into the indeterminate form inf/inf, so that I can apply L'Hopital's rule. I got 0 too.
0
13 years ago
#8
(Original post by dvs)
The way I did it was using the substitution t=1/xÂ² to transform it into the indeterminate form inf/inf, so that I can apply L'Hopital's rule. I got 0 too.
Yes that seems reasonable or you could point out that e^(1/x^2) > 1/x^2

and so e^(-1/x^2) < x^2
0
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